The following theorem, I feel, is not very well-known, though it is a particularly useful one for solving certain types of “limit” problems. Let me pose a couple of elementary problems and offer their solutions. First, the theorem.

Stolz-Cesàro: Let (a_n)_{n \ge 1} and (b_n)_{n \ge 1} be two sequences of real numbers, such that (b_n) is positive, strictly increasing and unbounded. Then,

\displaystyle \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{a_{n+1} - a_n}{b_{n+1} - b_n},

if the limit on the right hand side exists.

The proof involves the usual \epsilon - \delta method, and I will avoid presenting it here since it isn’t particularly interesting. Just as Abel’s lemma is the discrete analogue of integration by parts, the Stolz-Cesàro theorem may be considered the discrete analogue of L’Hospital’s rule in calculus.

Problem 1: Evaluate the limit \displaystyle \lim_{n \to \infty} \frac{1^k + 2^k + \ldots + n^k}{n^{k+1}}, where k \in \mathbb{N}.

Solution: One may certainly consider the above limit as a Riemann-sum which may then be transformed into the integral \displaystyle \int_0^1 x^k \, dx, which then obviously evaluates to 1/(k+1). But, we will take a different route here.

First, let a_n = 1^k + 2^k + \ldots + n^k and b_n = n^{k+1}. Then, we note that the sequence (b_n) is positive, strictly increasing and unbounded. Now,

\displaystyle \lim_{n \to \infty} \frac{a_{n+1} - a_n}{b_{n+1} - b_n} = \lim_{n \to \infty} \frac{(n+1)^k}{(n+1)^{k+1} - n^{k+1}}

\displaystyle = \lim_{n \to \infty} \frac{(n+1)^k}{\left(1 + \binom{k+1}{1}n + \binom{k+1}{2}n^2 + \ldots + \binom{k+1}{k}n^k + n^{k+1}\right) - n^{k+1}}

(using the binomial theorem)

\displaystyle = \lim_{n \to \infty} \frac{(n+1)^k / n^k}{\left(1 + \binom{k+1}{1}n + \binom{k+1}{2}n^2 + \ldots + \binom{k+1}{k}n^k \right) / n^k}

\displaystyle = \lim_{n \to \infty} \frac{(1 + 1/n)^k}{\binom{k+1}{k}} = \frac1{k+1}.

Therefore, using the Stolz-Cesàro theorem, we conclude that the required limit is also 1/(k+1).

Let us now look at another problem where applying the aforesaid theorem makes our job a lot easier. This problem is an example of one that is not amenable to the other usual methods of evaluating limits.

Problem 2: Let k\geq 2 be integers and suppose C: y = \sqrt {2x + 1}\ (x > 0). Given the tangent line at the point (a_{k},\, \sqrt {2a_{k} + 1}) from the point (0, k) to C, evaluate

\displaystyle \lim_{n\to\infty}\frac {1}{n^{3}}\sum_{k = 2}^{n}a_{k}.

Solution:(This is basically the solution I had offered elsewhere a while ago; so, it’s pretty much copy/paste!)

\displaystyle y = \sqrt {2x + 1}\Rightarrow \frac {dy}{dx} = \frac {1}{\sqrt {2x + 1}}.

So, the equation of the tangent line at the point (a_{k}, \sqrt {2a_{k} + 1}) is given by

\displaystyle y - \sqrt {2a_{k} + 1} = \frac {1}{\sqrt {2a_{k} + 1}}(x - a_{k}).

Since the point (0,k) lies on this line, we must have

\displaystyle k - \sqrt {2a_{k} + 1} = \frac {1}{\sqrt {2a_{k} + 1}}( - a_{k})

The above, after squaring and some algebraic manipulation yields
a_{k}^{2} + 2(1 - k^{2})a_{k} + 1 - k^{2} = 0, which implies a_{k} = (k^{2} - 1) + k(\sqrt {k^{2} - 1}). We drop the negative root because a_{k} > 0 for all k\ge 2.

(This is where the Stolz-Cesàro theorem actually comes into play!)

Now, let (b_{n}) and (c_{n}) be two sequences such that \displaystyle b_{n} = \sum_{k = 2}^{n}a_{k} and c_{n} = n^{3}.
Note that (c_{n}) is a positive, increasing and unbounded sequence.

Therefore, \displaystyle \lim_{n\to \infty}\frac {b_{n + 1} - b_{n}}{c_{n + 1} - c_{n}} = \lim_{n\rightarrow \infty}\frac {\sum_{k = 2}^{n + 1}a_{k} - \sum_{k = 2}^{n}a_{k}}{(n + 1)^{3} - n^{3}} = \lim_{n\rightarrow \infty}\frac {a_{n + 1}}{3n^{2} + 3n + 1}

\displaystyle = \lim_{n\to \infty}\frac {(n + 1)^{2} - 1 + (n + 1)\sqrt {(n + 1)^{2} - 1}}{3n^{2} + 3n + 1}

\displaystyle = \lim_{n \to \infty} \frac{(1 + 1/n)^2 - 1/n^2 + (1 + 1/n)\sqrt{1 + 2/n}}{3 + 3/n + 1/n^2}

= 2/3.

Therefore, by the Stolz- Cesàro theorem, we have

\displaystyle \lim_{n\to \infty}\frac {b_{n}}{c_{n}} = 2/3 \,, and so

\displaystyle \lim_{n\to \infty}\frac {1}{n^{3}}\sum_{k = 2}^{n}a_{k} = 2/3.